# Intersection of nested compact sets [closed]

Consider a topological space $$(X,\mathcal{T})$$, and a sequence of sets $$(K_n)_{n\in\omega}$$, such that $$(\forall n\in\omega)(K_n\text{ is compact})$$, and $$(\forall m. Is it true that $$\cap_{n\in\omega}K_n\neq\emptyset$$?

As pointed out elsewhere you need to assume the $$K_n$$ are all non-empty. However, that assumption is not strong enough: under the cofinite topology on $$\Bbb{R}$$, every subset of $$\Bbb{R}$$ is compact. So if you take $$K_n = \{n, n + 1, n + 2, \ldots\}$$, you get a counter-example to your claim.
If you assume $$(X, {\cal T})\,$$ is Hausdorff, your claim holds. (If $$\bigcap_n K_n\,$$ were empty, the sets $$K_1 \setminus K_n\,$$ would be open, because of Hausdorffness, and would cover $$K_1$$. Hence a finite subset of these open sets would cover $$K_1$$, implying that the $$K_n$$ are eventually all empty).
All the above argument relies on is that the $$K_i$$ are all closed. So it works if we take that as an assumption or if we assume that $$X$$ is a $$KC$$-space, one in which all compact subsets are closed.
• @RobArthan the point is, if $K_m = \{\}$ then you can't possibly have $K_m \supsetneq K_n$. Sep 26 at 8:31
• That's a good point, but the result is stronger to assume $\subset$ allows equality, with the caveat that sets are nonempty. Sep 26 at 11:16
• Perhaps you should mention in your answer that spaces in which all compact sets are closed are known as KC-spaces. And of course the claim is true if all $K_n$ are assumed to be closed ... Sep 29 at 15:44